Thanks Maureen,
I don't get the errors I was getting so that shows me you are great. Now my
next problem is if an item is found how do I echo or print that to the
screen. I have this as a starting point but I don't know where to go after
this.
if($search)
{
$query = "SELECT uid, id, image, iname, quantity, user, type FROM
{$config["prefix"]}_shop WHERE iname LIKE '%$searchword%' ORDER BY iname";
$ret = mysql_query($query) or die(mysql_error());
while(list($iname)=mysql_fetch_row($ret))
$user = $row['user'];
if($iname == $searchword) echo $user;
}else{
echo "<FORM METHOD=post ACTION='$PHP_SELF'>";
echo "Search Shops<BR><input type=text name=\"searchword\" value=\"\">";
echo "<BR><INPUT TYPE='submit' NAME='search' VALUE='Search'>";
}
I can't get any data to show up.
Could you shed some light on this. I don't want you to wirte it just let me
know if I am on the righ track.
Thanks
Jennifer
"Maureen" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> The query should read as below, without the = after the LIKE:
>
> $query = "SELECT uid, id, image, iname, quantity, type FROM
> {$config["prefix"]}_shop WHERE iname LIKE '%$shopsearch% ORDER BY
> iname'";
>
> HTH
>
> Maureen
>
>
> Jennifer Downey <[EMAIL PROTECTED]> said:
>
> > Hi all,
> >
> > I have been trying to build a search script for my site that deals with
only
> > one table in my db.
> > As Julie Meloni pointed out look in the MySQL manual for LIKE clauses I
> > can't seem to locate that clause in ether
> > manual.
> > Dan Brunner gave me this to go on:
> >
> > $query = "SELECT uid, id, image, iname, quantity, type FROM
> > {$config["prefix"]}_shop WHERE iname LIKE = '%$shopsearch% ORDER BY
> > iname'";
> > $ret = mysql_query($query);
> > while(list($quantity)=mysql_fetch_row($ret))
> >
> > But I can't seem to get this to work. I have never worked with LIKE
before
> > and would appreciate any help on this.
> >
> > Thanks Julie and Dan for your time and effort.
> >
> > Thanks all for your time and help
> > Jennifer
> >
> >
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